Hi, I have following optimization problem:
Min: x1 + x2 +...+ x7
subject to:
x1 + x2 = 80
x2 + x3 = 65
x3 + x4 = 40
all xi are ***positive integer***.
Can somebody help me in this optimization problem?
Thanks for your help
__
R-help@r-project.org
Dear all, I am looking for some procedure to apply 'ifelse' condition on
function. I have created an alternative to lapply() function with exactly same
set of arguments named lapply1(), however with different internal codes.
Therefore I want something like, if (some condition) then call
Dear all, while executing some function, there are some custom messages popping
up onto the R console and I do not want to see them. While looking into the
corresponding codes of those function, I see that those are coming from
message() function.
Is there any way to stop those messages coming
Let say i have a square matrix and applied the 'vech' operator to stack the
lower triangular elements into a vector:
Mat - matrix(1:25, 5)
Mat
[,1] [,2] [,3] [,4] [,5]
[1,] 1 6 11 16 21
[2,] 2 7 12 17 22
[3,] 3 8 13 18 23
[4,] 4 9 14 19 24
Let say, I have a character vector of arbitrary length:
Vector - c(a, b, c)
Using that vector I would like to create a matrix (with number of columns as 2)
with all pairwise combinations of those elements, like:
Vector - c(a, b, c)
Mat - rbind(c(a, b), c(a, c), c(b, c)); Mat # number of
Dear all, I have following kind of character vector:
Vec - c(344426, dwjjsgcj, 123sgdc, aagha123, sdh343asgh, 123jhd51)
Now I want to split each element of this vector according to numeric and string
element. For example in the 1st element of that vector, there is no string
element. Therefore
Dear all, I would like to draw a 3D plot as shown
here http://en.wikipedia.org/wiki/File:NaturalLogarithmAll.png, for this
function f = exp[ 1 - x^2 - y^2] (this function is some arbitrary!). I am
aware of different 3D plotting system in R, however it would be great if I can
get that kind of
Hi all, I am to find some way on how I can tell R to use this small number
10^-20 as zero by default. This means if any number is below this then that
should be treated as negative, or if I divide something by any number less than
that (in absolute term) then, Inf will be displayed etc.
I have
Please consider following string:
MyString - ABCFR34564IJVEOJC3434
Here you see that, there are 4 groups in above string. 1st and 3rd groups
are for english letters and 2nd and 4th for numeric. Given a string, how can
I separate out those 4 groups?
Thanks for your time
[[alternative
IJVEOJC 3434.
36453
Can you please tell me how can I modify that?
Thanks,
On Sun, Feb 13, 2011 at 11:10 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Sun, Feb 13, 2011 at 10:27 AM, Megh Dal megh700...@gmail.com wrote:
Please consider following string:
MyString
I need some help in defining a print method for my new S4 class
definition. So fer I have worked like this:
setClass(MyClass, sealed=F, representation(slot1 = list,#create a
new class
slot2= vector,
slot3 = vector,
slot4 = vector))
I am looking for an elegant way how I can test the equality of lengths of
multiple vectors.
For example, this is working fine:
length(rnorm(4)) == length(rnorm(5))
[1] FALSE
However this is not:
length(rnorm(4)) == length(rnorm(5)) == length(rnorm(6))
Error: unexpected '==' in
Dear all, I need to download an excel file from net, on which I have address
like http://www.2shared.com/file/MMSMWv4B/MyData.html. Can I somehow
directly download this file into my R workbook?
Thanks,
[[alternative HTML version deleted]]
__
Dear all, can somebody point me from where to download rcompgen package?
CRAN does not seem to hold that.
Installing this package through install.packages() tells this package is not
available.
Thanks
[[alternative HTML version deleted]]
__
Hi there, can anyone tell me how to extract to values of a particular slot for
some S4 object? Let take following example:
library(fOptions)
val -GBSOption(TypeFlag = c, S = 60, X = 65, Time = 1/4, r = 0.08, b =
0.08, sigma = 0.30)
val
Title:
Black Scholes Option Valuation
Call:
Dear friend, I have to construct some recursive algorithm for which I used some
for loop like:
res - vector(length=1)
res[1] = 0
for (i in 2:(1+1)) res[i] - res[i-1]*some function
I have noticed that this is taking too much time. Is there any way to speed up
things?
Thanks,
Dear all, I have following 2 zoo objects. However when I try to merge those 2
objects into one, nothing is coming as intended. Please see below the objects
as well as the merged object:
dat11
V2 V3 V4 V5
2010-10-15 13:43:54 73.8 73.8 73.8 73.8
2010-10-15 13:44:15
: [R] Problem with merging two zoo objects
To: Megh Dal megh700...@yahoo.com
Cc: r-h...@stat.math.ethz.ch
Date: Saturday, October 16, 2010, 12:11 AM
On Fri, Oct 15, 2010 at 2:20 PM, Megh
Dal megh700...@yahoo.com
wrote:
Dear all, I have following 2 zoo objects. However when
I try to merge
matter?
Thanks,
--- On Sat, 10/16/10, Megh Dal megh700...@yahoo.com wrote:
From: Megh Dal megh700...@yahoo.com
Subject: Re: [R] Problem with merging two zoo objects
To: Gabor Grothendieck ggrothendi...@gmail.com
Cc: r-help@r-project.org
Date: Saturday, October 16, 2010, 7:20 AM
I dont know
I dont know whether I am missing something or not:
head(read.zoo(file=f:/dat1.txt, header=T, sep=,, format = %m/%d/%Y
%H:%M:%S), tz=GMT)
data.open data.high data.low data.close
2010-10-15 73.7 73.7 73.7 73.7
2010-10-15 73.8 73.8 73.8 73.8
Can anyone please tell me how can use save.image() function if it is placed
within a function (i.e. some level up from the base level environment)? Here I
experimented with following codes:
#rm(list=ls())
fn - function() {
x - rnorm(5)
save.image(f:/dat.RData)
}
fn()
However I see
the same against
envir. By putting so, what I am going to tell R?
Thanks,
--- On Thu, 10/14/10, Joshua Wiley jwiley.ps...@gmail.com wrote:
From: Joshua Wiley jwiley.ps...@gmail.com
Subject: Re: [R] Query on save.image()
To: Megh Dal megh700...@yahoo.com
Cc: r-h...@stat.math.ethz.ch
Date: Thursday
Suppose I have following arbitrary matrix:
set.seed(1)
mat - matrix(rnorm(6), 3, 2)
mat
[,1] [,2]
[1,] -0.6264538 1.5952808
[2,] 0.1836433 0.3295078
[3,] -0.8356286 -0.8204684
Now I want to make a simple object like (character type):
Hi, I was trying to split the following matrix dat:
set.seed(1)
dat - matrix(rnorm(4*16), 4, 16)
dat
[,1] [,2] [,3][,4][,5][,6]
[,7] [,8][,9] [,10] [,11]
[1,] -0.6264538 0.3295078 0.5757814 -0.62124058 -0.01619026
Hi, I want to split a text to seperate numerical and non-numerical portions of
that. For example suppose I have a text abc 3456 and I want to split in 2
parts like abc 3456.
Is there any function to do that?
Thanks,
__
R-help@r-project.org mailing
Hi, is there any way to say: this class 'x' is a S3 class? For example what
is the type of class data.frame? Is it a S3 class or S4?
How can I get a complete list of all S3 classes currently available?
Thanks,
__
R-help@r-project.org mailing list
Here I am having problem to define a subclass, specially if I define that
subclass after defining initialize() method for its superclass. Here is my code:
setClass(a, representation=list(x=numeric, y=numeric),
prototype=list(x=rnorm(10), y=rnorm(10)))
[1] a
setMethod(initialize, a,
understanding,
all defined law for super-class should be inherited by it's sub-class,
therefore no need to define again.
I would be really grateful if someone clarify those.
Thanks
--- On Fri, 7/30/10, Megh Dal megh700...@yahoo.com wrote:
From: Megh Dal megh700...@yahoo.com
Subject: Having problem
Hi all, I have following environments loaded with my current R session:
search()
[1] .GlobalEnvpackage:stats package:graphics
package:grDevices
[5] package:utils package:datasets package:methods Autoloads
[9] package:base
How can I find the objects under a
: [R] Objects within environment
To: Megh Dal megh700...@yahoo.com
Cc: r-h...@stat.math.ethz.ch
Date: Wednesday, July 21, 2010, 4:48 PM
On 21/07/2010 5:57 AM, Megh Dal
wrote:
Hi all, I have following environments loaded with my
current R session:
search()
[1] .GlobalEnv
Thanks Duncan, I understood. Your explanation is really great. Thank you so
much for your time.
--- On Wed, 7/21/10, Duncan Murdoch murdoch.dun...@gmail.com wrote:
From: Duncan Murdoch murdoch.dun...@gmail.com
Subject: Re: [R] Objects within environment
To: Megh Dal megh700...@yahoo.com
Cc
Dear all, I need to download some data from this webpage:
http://markets.ft.com/ft/markets/researchArchive.asp
Notable thing here is that there are some fields to be selected to get the
desired data. Is there any R facility to do this directly?
Obviously I can do it manually and then just
Hi all, can somebody help me to split a time series (zoo) object on monthwise.
For example, suppose I have following time series object:
library(zoo)
dat1 - zooreg(rnorm(300), start=as.Date(2009-01-01), frequency=1)
From dat1, I want to create a list-object dat2 like:
dat2[[1]] - all
Dear all,
Please forgive me if there is a duplicate post; my previous mail perhaps didnt
reach the list...
Let say I have following time series
library(zoo)
dat1 - zooreg(rnorm(10), start=as.Date(2010-01-01), frequency=1)
dat1[c(3, 7,8)] = NA
dat1
2010-01-01 2010-01-02 2010-01-03
Hi all, good morning,
My question is not really R related rather a practical problem and wondering if
stat-gurus here can show some light how that can be solved with some
statistical/mathematical tool.
I have some 10 items on which 10,000 viewers put their views based on some 12
attributes
/12/10, Erik Iverson er...@ccbr.umn.edu wrote:
From: Erik Iverson er...@ccbr.umn.edu
Subject: Re: [R] Handling character string
To: Megh Dal megh700...@yahoo.com
Cc: r-h...@stat.math.ethz.ch
Date: Saturday, June 12, 2010, 2:36 AM
Megh Dal wrote:
Dear all, Is there any R function to say these 2
...@gmail.com
Subject: Re: [R] Handling character string
To: Megh Dal megh700...@yahoo.com
Cc: Erik Iverson er...@ccbr.umn.edu, r-h...@stat.math.ethz.ch
Date: Saturday, June 12, 2010, 10:18 PM
This is probably what you want:
sub(^[[:space:]]*, , Now is the time)
[1] Now is the time
You need
Dear all,
Is there any R function to say these 2 character strings temp and temp
are actually same? If I type following code R says there are indeed different :
temp == temp[1] FALSE
Is there any way out?
[[alternative HTML version deleted]]
Hi all, Here I am trying to implement the switch() function to choose value of
a variable depending on the value of an input variable :
temp1 - 1
temp1.name - switch(temp1,
1 == aa,
2 == bb,
Dear falks, here I have written following function :
fn - Vectorize(function(x = 1:3, y = 3:6) {
x - matrix(x, nrow=1)
y - matrix(y, ncol=1)
dat - apply(x, 2, function(xx) {
apply(y, 1, function(yy) {
Dear folks, I have created a plot on RGL device :
x = 1:6
y = seq(-12, 5, by=1)
z = matrix(0, length(y), length(x))
z[13,3] = 1; z[13,4] = 1.011765
surface3d(x, y, t(z), col=rainbow(1000))
grid3d(c(x-, y-, z))
Now I want to draw 2 lines along x=3 x=4, over the surface (with different
Murdoch murdoch.dun...@gmail.com
Subject: Re: [R] [RGL] Need help to modify current plot
To: Megh Dal megh700...@yahoo.com
Cc: r-h...@stat.math.ethz.ch
Date: Tuesday, May 18, 2010, 3:51 PM
Megh Dal wrote:
Dear folks, I have created a plot on RGL device :
x = 1:6
y = seq(-12, 5, by=1)
z
Hi all, previously I submitted this thread through Nabble which seems fail
therefore sending it again
suppose I have written following function :
fn = function(x) return(x+x^2)
fn
function(x) return(x+x^2)
Here you see, if I type only the function name all inside information of this
Can anyone please tell me how to define a list. Suppose I want to define a
list object result with length n then want to fill each place of result
with different objects. For e.g.
i=1
result[1] = rnorm(1)
i=2
result[2] = rnorm(2)
...
i=n
result[n] = rnorm(n)
What would
I have 2 vecros :
x-c(100,96,88,100,100,96,80,68,92,96,88,92,68,84,84,88,72,88,72,88)
x1 = sample(x, 5, replace=FALSE)
Now i want to get remaining values of vector x those are not member of vector
x1. Can anyone please tell me how to do that?
__
PROTECTED]
Cc: [EMAIL PROTECTED]
Date: Sunday, October 12, 2008, 1:06 AM
Hi Megh,
Try this:
x-c(100,96,88,100,100,96,80,68,92,96,88,92,68,84,84,88,72,88,72,88)
x1 = sample(x, 5, replace=FALSE)
x[ ! x %in% x1]
HTH,
Jorge
On Sat, Oct 11, 2008 at 3:25 PM, Megh Dal
[EMAIL PROTECTED
Hi,
I am trying to draw a Q-Q plot, however got following error.
library(sn)
library(car)
dat1 = rst(1000, 0, 1, 0, 2)
qq.plot(dat1, st, 0, 1, 0, 9)
Error in plot.window(...) :
invalid value specified for graphical parameter las
Can anyone please tell me why this error is coming?
I have drawn a 3D scatter plot :
library(mnormt)
library(scatterplot3d)
dat = cbind(rmnorm(3, rep(0,2), diag(2)), 1:3)
scatterplot3d(dat)
Now I want to do 2 things :
1 : In the Z-axis (i.e. height), I want to see only numbers 1,2,3, etc NOT,
1,1.5,2,2.5.
2. I want to add two
)
--- On Thu, 10/2/08, Uwe Ligges [EMAIL PROTECTED] wrote:
From: Uwe Ligges [EMAIL PROTECTED]
Subject: Re: [R] Adding plane in a 3D scatterplot
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Date: Thursday, October 2, 2008, 11:38 PM
Megh Dal wrote:
I have drawn a 3D scatter plot :
library
When I try to load fBasics package, I get following error/warning :
library(fBasics)
Loading required package: fImport
Loading required package: fSeries
Loading required package: fBasics
Loading required package: fImport
Loading required package: fSeries
Loading required package: fBasics
Loading
I want to generate a valid variance-covariance matrix. One way could be to
generate some random sample from multivariate normal distribution and then
calculate cov. matrix. Another way could be to sample from wishart distribution
itself. However both cases need a valid i.e. PD covariance
I have following matrix :
a = matrix(rnorm(36), 6)
Now I want to replace the lower-triangular elements with it's upper-triangular
elements. That is I want to make a symmetric matrix from a. I have tried with
lower.tri() and upper.tri() function, but got desired result. Can anyone please
tell
hope it helps.
Best,
Dimitris
Megh Dal wrote:
[My previous message rejected, therefore I am sending
same one with some modification]
I have 3 vectors with object name : dat1, dat2, dat3
Now I want to create a loop, like :
for (i in 1:3)
{
cat(sd(dati
)
lis
output - sapply(lis, sd)
print(output)
print(str(output))
On Fri, Sep 19, 2008 at 1:44 AM, Megh Dal wrote:
Thanks for this mail. It runs perfectly, but now I
stuck on how to
convert the result in a vector format for further
matrix-compputation.
e.g. How to convert sapply
] Compiling date
To: David Scott [EMAIL PROTECTED]
Cc: Dr Eberhard W Lisse [EMAIL PROTECTED], Megh Dal [EMAIL
PROTECTED], [EMAIL PROTECTED]
Date: Tuesday, September 9, 2008, 11:37 PM
Is this Month-Day or Day-Month or a mixture of both?
I still think using the Format - Cell - Date will
work
much
I have following
1975 01 7711.16
Here I need to identify where the space is there and then concatenate rest of
the digits without space, i.e. I want to have 1975017711.16. Is there any R
function?
Regards,
__
R-help@r-project.org mailing list
I have following dataset:
res
[,1] [,2] [,3]
[1,] 19464 1.27
[2,] 19465 1.27
[3,] 19466 1.27
[4,] 19467 1.27
[5,] 19468 1.52
[6,] 19469 1.52
[7,] 1946 10 1.52
[8,] 1946 11 1.52
[9,] 1946 12 1.62
[10,] 19471 1.62
[11,] 19472 1.62
[12,] 1947
Hi,
I have following kind of dataset (all are dates) in my Excel sheet.
09/08/08
09/05/08
09/04/08
09/02/08
09/01/08
29/08/2008
28/08/2008
27/08/2008
26/08/2008
25/08/2008
22/08/2008
21/08/2008
20/08/2008
18/08/2008
14/08/2008
13/08/2008
08/12/08
08/11/08
08/08/08
08/07/08
However I want to use
I have created following interactive plot :
library(TeachingDemos)
plott = function(x)
{
return(hist(rnorm(as.integer(1000, 10, x)), xlab=NA))
}
tkexamp(plott, list(x=list('slider',from=1,to=40, resolution=0.1, init=2)),
plotloc='top')
Here everything works fine, but the problaem is
: Thursday, July 24, 2008, 5:42 PM
On 7/24/2008 7:34 AM, Megh Dal wrote:
I have created following interactive plot :
library(TeachingDemos)
plott = function(x)
{
return(hist(rnorm(as.integer(1000, 10, x)),
xlab=NA))
}
tkexamp(plott,
list(x=list('slider',from=1,to=40
I used ggplot to create a scatter plot :
library(ggplot)
library(mnormt)
Sigma = matrix(c(1, 0.6, 0.6, 1), 2, 2)
x = rmnorm(20, c(0,0), Sigma)
xx = x[order(x[,1]),]
y = xx[,1]
z = xx[,2]
qplot(z, y, type=point, main=x-y plot, xlab=x, col=blue)
However I want following:
1. Plot color must be
Hi,
following data is taken from
http://www.economagic.com/em-cgi/data.exe/var/west-texas-crude-long. Problem
with this data is when you copy it from this site you would get something like
that :
1946 063331.27
1946 079991.27
1946 087771.52
1946 096661.52
They should be interpret in
, sapply(tolower(funs), f, simplify = FALSE))
On Sun, Jul 6, 2008 at 10:43 PM, Megh Dal
[EMAIL PROTECTED] wrote:
I made some changes and also incorporated your advice
:
library(zoo)
Z.index - as.Date(sample(12450:15500, 3000))
Z.data - matrix(rnorm(300), ncol = 1)
data1 - zoo
Can anyone please tell me why I am getting this error?
library(zoo)
Z.index - as.Date(sample(12450:15500, 3000))
Z.data - matrix(rnorm(300), ncol = 1)
data1 - zoo(Z.data, Z.index)
fnc = function(data1)
{
selection2 = select.list(c(Mean), multiple = F)
Mean = function(dataa)
in defining function
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Date: Monday, July 7, 2008, 1:23 AM
On Sun, Jul 6, 2008 at 3:19 PM, Megh Dal
[EMAIL PROTECTED] wrote:
Can anyone please tell me why I am getting this error?
library(zoo)
Z.index - as.Date(sample(12450:15500, 3000))
Z.data
I have time series observation on daily frequencies :
library(zoo)
SD=1
date1 = seq(as.Date(01/01/01, format = %m/%d/%y), as.Date(12/31/02,
format = %m/%d/%y), by = 1)
len1 = length(date1); data1 = zoo(matrix(rnorm(len1, mean=0, sd=SD*0.5), nrow =
len1), date1)
plot(data1)
Now I want to
Hi,
I have one question on expand.grid() function.
When I write following syntax :expand.grid(c(u, l), c(u, l), c(u,
l)) I get following as desired :
Var1 Var2 Var3
1uuu
2luu
3ulu
4llu
5uul
6lul
7ul
Hi,
I got following error in write.table() :
write.table(dataa, file=c:/data1.csv, row.names=F, col.names=T, sep=,)
Error in file(file, ifelse(append, a, w)) :
cannot open the connection
In addition: Warning message:
In file(file, ifelse(append, a, w)) :
cannot open file 'c:/data1.csv':
Hi all,
I have following monthly time series :
head(data1)
V1 V2 V3
1 Nov-80 NA 1007.44
2 Dec-80 NA 982.05
3 Jan-81 NA 994.25
4 Feb-81 NA 996.31
5 Mar-81 NA 939.91
6 Apr-81 NA 923.32
Now I want to convert it to a 'zoo' object. I wrote following syntax :
ss = zoo(data1[,3],
-01 0005-01-01 0006-01-01
1206.68 782.451187.001398.771883.231431.80
Nowhere 1980 is coming. Any better suggestion?
Gabor Grothendieck [EMAIL PROTECTED] wrote: On Sun, Jun 1, 2008 at 8:47 AM,
Gabor Grothendieck
wrote:
On Sun, Jun 1, 2008 at 8:35 AM, Megh Dal wrote:
Hi
, 2008 at 9:37 AM, Megh Dal wrote:
I got following:
z - zoo(data1[,3], as.yearmon(data1[,1], %b-%y))
head(z)
Jan 0001 Jan 0002 Jan 0003 Jan 0004 Jan 0005 Jan 0006
1206.68 782.45 1187.00 1398.77 1883.23 1431.80
z - zoo(data1[,3], as.Date(as.yearmon(data1[,1], %b-%y)))
head(z)
0001-01
provide the dput output as
requested.
On Sun, Jun 1, 2008 at 9:59 AM, Megh Dal wrote:
packageDescription(zoo)$Version
[1] 1.1-1
Gabor Grothendieck wrote:
You need to unambiguously specify what your data
looks like. Please provide the dput output as previously
requested. Also what version
Still I did not find any suggestion. Is my problem not elaborate enough?
Megh Dal [EMAIL PROTECTED] wrote: I think I should be clear exactly what I
want :
take following example :
a = b = seq(1, 5, by=500)
v = matrix(0, nrow=length(a), ncol=length(a))
for (i in 1:length
I got following error while I was using solve.QP() in my problem:
Dmat = matrix(c(0.0001741, 0.0001280, 0.0001280, 0.0002570), nrow=2)
dvec = t(c(0,0))
Amat = matrix(c(-1,1,0,-1,0, 1,0,1,0,-1), nrow=5)
bvec = c(-2, 1, 1, -5, -5)
solve.QP(Dmat,dvec,Amat,bvec=bvec)
Error in
Hi,
I want to find solution of function : f(x,y) = x'Cx - a under constraints :
0 x,y p
0 x-y q
where a, p,q are given constants and x = (x, y) and C is a 2X2 matrix (given)
Can anyone suggest me any R function to do that?
Forgot to send one copy to R help. Sorry
Megh Dal [EMAIL PROTECTED] wrote: Date: Wed, 7 May 2008 02:45:09 -0700 (PDT)
From: Megh Dal [EMAIL PROTECTED]
Subject: Re: [R] Solution of function
To: Berwin A Turlach [EMAIL PROTECTED]
Hi Berwin,
Thanks for having look on my problem. However
is the corresponding
component values of x.
Any suggestion please?
Peter Dalgaard [EMAIL PROTECTED] wrote: Megh Dal wrote:
Hi,
I want to find solution of function : f(x,y) = x'Cx - a under constraints :
0 x,y p
0 x-y q
where a, p,q are given constants and x = (x, y) and C
Hi all,
I have following problem :
a = b = seq(1, 5, by=500)
v = matrix(0, nrow=length(a), ncol=length(a))
for (i in 1:length(a))
{
for (j in 1:length(a))
{
d = c(17989*a[i], -18109*b[j])
v[i,j] = t(d) %*% matrix(c(0.0001741, 0.0001280, 0.0001280,
I have a TS of monthly observations.
head(data4)
1991(1) 1991(2) 1991(3) 1991(4) 1991(5) 1991(6)
12.00864 11.94203 11.98386 12.01900 12.19226 12.15488
Now I want to make 11 dummy variables indicating months. Therefore I did
followings :
For Jan :
rep(c(rep(0,0), 1,
Hi all,
I feel there is a incompleteness in mAr.est function in mAr package for VAR
estimation. I does not cheak whether there is multicolinearity in data set.
Here I used mAr.est function for following dataset :
head(log(data1)
+ )
V1 V2 V3 V4 V5
) etc
However I can not reproduce this example in my problem. How I can change my
data matrix 'dat' to 'Z' ?
If anyone show me some light it would be great for me.
Regards,
Megh Dal [EMAIL PROTECTED] wrote:
Roy Mendelssohn [EMAIL PROTECTED] wrote: Date: Thu, 28 Feb 2008 20
tried with image() function [however there is also no option to
put a color pallet :( ], there is a option 'zlim' to trim the not-required
values. Is there any option in levelplot() function as well? I already gone
through it's help file, however got nothing on that
Megh Dal [EMAIL PROTECTED
Hi all,
Suppose I have following dataset :
library(zoo)
SD = 1
date1 = seq(as.Date(01/01/90, format = %m/%d/%y), as.Date(12/31/08,
format = %m/%d/%y), by = 1)
len1 = length(date1); data1 = zoo(matrix(rnorm(len1, mean=0, sd=SD*0.5), nrow =
len1), date1)
Now I want to extract
Suppose I have following dataset :
head(data1)
Date Return
1 03/31/00 0.14230650
2 04/28/00 -0.03276228
3 05/31/00 -0.06527890
4 06/30/00 -0.04999873
5 07/31/00 -0.01447902
6 08/31/00 0.22265729
Now I convert it to zoo object :
data11 = zoo(data1[,2], as.Date(data1[,1],
Hi all,
Suppose I have to letters 'u' and 'd'. Now I want to find all combinations
like that :
uu
ud
du
.
dd
This type of combination generally required for Option Pricing in
financial/derivative market. Here generally 'u' means: up-move and 'd' means
down
tell me how to do that, it will be good for
me.
Regards,
Megh Dal [EMAIL PROTECTED] wrote: Hi all,
Can anyone here please tell me whether is it possible to produce a chart
displayed in http://www.datawolf.blogspot.com/ in R for visualizing
multivariate time series? If possible how?
Regards
)
r - sqrt(outer(x^2, y^2, +))
image(x, y, r, col=gray((0:32)/32))
colors - colorRampPalette(c('red', 'yellow', 'blue')) # create you
color spectrum
image(x,y,r, col=colors(100))
On Thu, Feb 28, 2008 at 9:28 PM, Megh Dal wrote:
I used ?image function to do that, like below :
require(grDevices
let create a 'zoo' object :
library(zoo)
date.data = seq(as.Date(01/01/01, format = %m/%d/%y), as.Date(06/25/02,
format = %m/%d/%y), by = 1)
len = length(date.data)
data1 = zoo(matrix(rnorm(2*len), nrow = len), date.data )
head(data1)
Now I want to create an 3 dimensional
this:
lapply(split(data1, format(index(data1), %m)), cov)
On 27/02/2008, Megh Dal wrote:
let create a 'zoo' object :
library(zoo)
date.data = seq(as.Date(01/01/01, format = %m/%d/%y), as.Date(06/25/02,
format = %m/%d/%y), by = 1)
len = length(date.data)
data1 = zoo(matrix(rnorm(2*len), nrow
Hi, I would like to ask here one stat related question. Suppose Z ~
Skew-Normal(0,1,1). Now I want to find a variable f: Z - Y which has Z ~
Skew-Normal(0,1,lambda) distribution. Can anyone give me some light how to do
that?
Regards,
-
Hi, I would like to ask here one stat related question. Suppose Z ~
Skew-Normal(0,1,1). Now I want to find a variable f: Z - Y which has Y~
Skew-Normal(0,1,lambda) distribution. Can anyone give me some light how to do
that?
Regards,
-
Hi all,
Can anyone here please tell me whether is it possible to produce a chart
displayed in http://www.datawolf.blogspot.com/ in R for visualizing
multivariate time series? If possible how?
Regards,
-
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-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Attiglah, Mama
Sent: 30 January 2008 11:32
To: Megh Dal; [EMAIL PROTECTED]
Subject: Re: [R] Multiplying each row of a big matrix with a vector
Ret
Hi all T gurus,
I would like to test if my dataset is indeed from N(0, 0.011908969).
K.S. test gives following result:
ks.test(data, pnorm, 0, 0.011908969)
One-sample Kolmogorov-Smirnov test
data: data
D = 0.1092, p-value = 1.318e-05
alternative hypothesis:
I have a big matrix 'ret'. I want to multiply each row of it with a 2nd vector
'pos', resulting result, I want to save in a vector named 'port'. I wrote
following code:
pos
[1] 2593419 2130220 6198197 1673888 198 1784732 2052120 -7490228
-5275000
dim(ret)
[1] 500 9
Hi all,
Suppose I have a population of 3 alphabets : A, B, C. From this population,
number of ways that any 2 can be chosen is 3 i.e. AB, AC, and BC.
Is there any R function to generalize this process, for any number of
alphabets/numbers and for any sub-sample size?
Thanks and
Hi all,
I have a time series like that :
DateValue
01/03/05 -0.008471364
01/04/05 -0.008153802
01/05/05 -0.000780031
01/06/05 -0.000130064
01/07/05 -0.000650576
01/08/05 -0.000130166
01/10/05 -0.004174282
01/11/05 0.01027384
01/12/05 -0.006099558
01/13/05
Hi all,
I have a time series like that :
DateValue
01/03/05 -0.008471364
01/04/05 -0.008153802
01/05/05 -0.000780031
01/06/05 -0.000130064
01/07/05 -0.000650576
01/08/05 -0.000130166
01/10/05 -0.004174282
01/11/05 0.01027384
01/12/05 -0.006099558
01/13/05
Hi all,
I have a time series like that :
DateValue
01/03/05 -0.008471364
01/04/05 -0.008153802
01/05/05 -0.000780031
01/06/05 -0.000130064
01/07/05 -0.000650576
01/08/05 -0.000130166
01/10/05 -0.004174282
01/11/05 0.01027384
01/12/05 -0.006099558
01/13/05
Hi all,
I have a time series like that :
DateValue
01/03/05 -0.008471364
01/04/05 -0.008153802
01/05/05 -0.000780031
01/06/05 -0.000130064
01/07/05 -0.000650576
01/08/05 -0.000130166
01/10/05 -0.004174282
01/11/05 0.01027384
01/12/05 -0.006099558
01/13/05
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